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Q. If $f\left(x\right) = \frac{2- x\cos x}{2+x \cos x}$ and $ g\left(x\right) =\log_{e}x ., \left(x>0\right) $ then the value of integral $\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}} g\left(f\left(x\right)\right)dx $ is :

JEE MainJEE Main 2019Integrals

Solution:

$g\left(f\left(x\right)\right) =\ell n\left(f\left(x\right)\right) =\ell n\left(\frac{2-x.\cos x}{2+x.\cos x}\right) $
$ \therefore I = \int^{\frac{\pi}{4}}_{-\frac{\pi}{4}} \ell n\left(\frac{2-x.\cos x}{2+x\cos x}\right)dx $
$ = \int^{\frac{\pi}{4}}_{0} \left(\ell n \left(\frac{2-x\cos x}{2+x.\cos x}\right) + \ell n\left(\frac{2+x.\cos x}{2-x.\cos x}\right)\right)dx $
$ = \int^{\frac{\pi}{2}}_{0} \left(0\right)dx =0 =\log_{e} \left(1\right) $